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Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$

Doesn't this thing approaches 0 at the end? why does it approaches 1?

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  • $\begingroup$ would you mind to write out the product series in a formula? $\endgroup$ – al-Hwarizmi Aug 4 '13 at 15:44
  • $\begingroup$ sorry i couldnt post the question on my account due to the stupid 24 hour limit $\endgroup$ – Commander Shepard Aug 4 '13 at 15:46
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    $\begingroup$ The last factor in the product, $3^{3^{-n}}$ doesn't go to 0, it approaches 1. $\endgroup$ – Francis Adams Aug 4 '13 at 15:51
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    $\begingroup$ How could it possibly be zero, given that each term is bigger than $1$? $\endgroup$ – Thomas Andrews Aug 5 '13 at 0:20
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Hint: $3^9\cdot3^3\cdot3^1\cdot\dots=3^{9+3+1+\cdots}$

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HINT:

Using Exponent Combination Laws, $$a^m\cdot a^n\cdot a^p\cdots=a^{m+n+p+\cdot},$$

$$\displaystyle 3^9\cdot 3^3\cdot3\cdot 3^\frac13\cdots=3^{\left(3^2+3+1+\frac13+\cdots\right)}$$

Observe that the power of $3$ is an infinite Geometric Series with the first Term $=9$ and common ratio $=\frac13<1$

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$$3^9 * 3^3 * 3 * 3^{\frac{1}{3}} * ...=$$

$$3^{9\sum_{n=0}^{\infty}3^{-n}}=$$ $$3^{9*1.5}=$$ $$3^{13.5}$$

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$ 3^{12} \times 3^{sum\ of\ geometric\ series }$

geometric series is

$ 1 + 1/3 + 1/9 + .... $

$ = 1 / (1-1/3) $

$ = 3/2 $

so

$= 3^ {12 + \frac{3}{2} } $

$= 3 ^{ 27/2 } $

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